Asked by Danette
Find the smallest N such that N!is evenly divisable by 12^19.
Answers
Answered by
Reiny
12^19 in prime factored form is
1^18 * 2^19 * 2^19 * 3^19 because 12 = 1x2x2x3
= 1^19 * 2 * 4 * 2^16 * 3 * 3^18
= 1x2x3x4 = 4!
since there is no factor of 5, that is as high as we can go
so N = 4
(4! divides into 12^19)
1^18 * 2^19 * 2^19 * 3^19 because 12 = 1x2x2x3
= 1^19 * 2 * 4 * 2^16 * 3 * 3^18
= 1x2x3x4 = 4!
since there is no factor of 5, that is as high as we can go
so N = 4
(4! divides into 12^19)
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